Hai bình kín ................
Giải thích
\({v_c} = \sqrt {\frac{{3kT}}{m}} \Rightarrow \frac{{{v_{cA}}}}{{{v_{cB}}}} = \sqrt {\frac{{{m_B}}}{{{m_A}}}} = 2 \Rightarrow \frac{{{m_B}}}{{{m_A}}} = 4.{\rm{ }}\)Chọn A
\({v_c} = \sqrt {\frac{{3kT}}{m}} \Rightarrow \frac{{{v_{cA}}}}{{{v_{cB}}}} = \sqrt {\frac{{{m_B}}}{{{m_A}}}} = 2 \Rightarrow \frac{{{m_B}}}{{{m_A}}} = 4.{\rm{ }}\)Chọn A